我使用递归创建了一个简单的脚本,以改变图像的位置。
它遍历它们并调整它们的位置,脚本是相对简单的代码:
var img = document.getElementsByClassName("container")[0];
var X = 0;
var Y = 0;
var check;
function all() {
var decko = img.children;
for (var i = 0; i < decko.length; i++) {
if (i !== 0 && i % 3 === 0) {
X = 0;
Y++;
}
decko[i].style.transform = "translate3d(" + X * 100 + "px," + Y * 100 + "px,0)";
X++;
}
X = 0;
Y = 0;
check = false;
}
all()
window.onclick = function() {
pushIt(img.length, img.children, 0, 0)
}
function pushIt(max, target, index, count) {
if (count == max) {
return;
}
var tmp = target[index];
var matrix =window.getComputedStyle(tmp).getPropertyValue("transform");
var translate_left=matrix.split(",")[4];
var translate_top=matrix.split(",")[5].split(")")[0]-100;
tmp.style.transform = "translate3d(" + translate_left + "px," + translate_top + "px,0)";
setTimeout(function(){
pushIt(max, target, index + 1, count + 1);
},150)
}
Here你可以看到它是如何运作的。问题是当有很多图像时。
当我点击调用该函数时,它会遍历所有图像(在我的情况下为30 +)。如果我在一秒钟内单击两次,它将循环遍历所有图像,然后它将执行第二次功能,这在我的情况下是不需要的行为(看起来很迟钝)。有什么方法可以防止这种行为吗?我选择了递归,bcs这似乎是最简单的选择。
答案 0 :(得分:1)
如果您的JavaScript出现了一些错误,那么您有一对错误,即您的函数中未定义max。我有这个工作HERE
var img = document.getElementsByClassName("container")[0];
var decko = img.children;
var X = 0;
var Y = 0;
var check;
var running = false;
function all() {
for (var i = 0; i < decko.length; i++) {
if (i !== 0 && i % 3 === 0) {
X = 0;
Y++;
}
decko[i].style.transform = "translate3d(" + X * 100 + "px," + Y * 100 + "px,0)";
X++;
}
X = 0;
Y = 0;
check = false;
}
all()
window.onclick = function () {
if (!running) {
running = true;
pushIt(decko.length, img.children, 0, 0);
}
}
var pushIt = function (max, target, index, count) {
if (count == max) {
running = false;
return;
}
var tmp = target[index];
var matrix = window.getComputedStyle(tmp).getPropertyValue("transform");
var translate_left = matrix.split(",")[4];
var translate_top = matrix.split(",")[5].split(")")[0] - 100;
tmp.style.transform = "translate3d(" + translate_left + "px," + translate_top + "px,0)";
setTimeout(function () {
console.log(running);
pushIt(max, target, index + 1, count + 1);
}, 150)
}